The international space station makes 15.65 revolutions per day in its orbit around the earth. part a part complete assuming a circular orbit, how high is this satellite above the surface of the earth

Considering that 
,
the frequency of revolution of the international space spation around Earth is

and so its orbital period is


Now we can use Kepler's third law to find the radius of the orbit of the space station, r. Knowning that the Earth mass is , Kepler's third law can be written as

where G is the gravitational constant. Re-arranging the formula and using the value of T we found previously, we find the orbital radius:

which corresponds to r=6750 km.

This is the radius of the orbit measured from the center of Earth. To find the distance of the space station from the Earth's surface, we must subtract the Earth's radius, R=6370 km, and we find:


So, the space station is approximately 380 km above Earth's surface.

What formula  this one  (x_/